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Inequality 25(Generalization)

The following problem is a generalization of a well known inequality:


Let \displaystyle a,b,c be positive real numbers such that \displaystyle ab+bc+ca=3. Prove that

\displaystyle \frac{1}{(a+kb)^3}+\frac{1}{(b+kc)^3}+\frac{1}{(c+ka)^3}\geq \frac{3}{(k+1)^3}, where \displaystyle k is a non-negative real number.


From Holder’s inequality we get that

\displaystyle \left[\sum_{cyc}\frac{1}{(a+kb)^3}\right]\left[\sum_{cyc}a(b+kc)\right]^{3}\geq \left(\sum_{cyc}a^{\frac{3}{4}}\right)^{4}, or \displaystyle \sum_{cyc}\frac{1}{(a+kb)^3}\geq \frac{\left(a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}\right)^{4}}{(k+1)^3(ab+bc+ca)^3}.

So, it suffices to prove that

\displaystyle \frac{\left(a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}\right)^{4}}{(k+1)^3(ab+bc+ca)^3}\geq \frac{3}{(k+1)^3}.

But the last relation is equivalent to \displaystyle \left(\sum_{cyc}a^{\frac{3}{4}}\right)^{4}\geq 81=9\sqrt{3}(ab+bc+ca)^{3/2}.

Let us denote by \displaystyle x,y,z the \displaystyle a^{3/4},b^{3/4},c^{3/4} respectively. Then \displaystyle ab=(xy)^{4/3},bc=(yz)^{4/3},ca=(zx)^{4/3}.

So, our inequality takes the form \displaystyle (x+y+z)^4\geq 9\sqrt{3}\sum_{cyc}(xy)^{4/3}. This inequality is homogeneous so, we consider the sum \displaystyle x+y+z equal to \displaystyle 3.

Doing some manipulations in left and right hand side we only need to prove that

\displaystyle \sum_{cyc}(xy)^{4/3}\leq 3.

Now from the AM-GM inequality we have that

\displaystyle \sum_{cyc}xy\cdot (xy)^{1/3}\leq \sum_{cyc}xy\frac{x+y+1}{3}=\sum_{cyc}xy\frac{4-z}{3}.

The last one is equal to \displaystyle \sum_{cyc}(xy)^{4/3}\leq \frac{4}{3}\sum_{cyc}xy-xyz.

After that, we only need to prove

\displaystyle \frac{4}{3}\sum_{cyc}xy-xyz\leq 3\Longrightarrow 4\sum_{cyc}x\sum_{cyc}xy\leq 27+9xyz=\left(\sum_{cyc}x\right)^{3}+9xyz,

which is Schur’s 3rd degree inequality, Q.E.D.


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